In discussion of LHC searches for SUSY particles, physicists seem to assume they will decay quickly to the lightest SUSY particle which then remains stable (at least within the time it takes to leave the detector). As an example: http://physics.aps.org/articles/v4/27

The stability of the lightest SUSY particle is desired as it provides an explanation of dark matter. This stability is usually given by postulating R-parity, the conservation of super-symmetric particles (kind of like lepton, or baryon number).

Hopefully my understanding up to this point is mostly correct. My question is then in two parts:

But if we assume R-parity, then wouldn't the lightest electrically charged SUSY particle also be stable? However we don't see charged dark matter.

So if it can decay, into the lightest SUSY particle + some standard model particles to get rid of the charge, AND theorists expect this decay to be so very rapid that only the decay products will be visible to detectors (so "high cross section" for decay reaction), THEN wouldn't the reverse reaction also have to have a huge cross section? That is, if colliding (SM stuff) + (dark matter) --> (other SM stuff) + (charged SUSY particle), has such a HUGE cross section, and dark matter is everywhere, then we don't need the energy to create a SUSY particle to see SUSY particle, we only need roughly the energy difference between lightest SUSY and lightest charged SUSY particle. In which case, if the cross section is really that big, it seems strange we haven't already seen this.

  • $\begingroup$ I"m pretty sure that the answer to this is that, if SUSY exists, the lightest SUSY particle is uncharged, and that this is an argument that SUSY models that predict that the lightest SUSY particle is charged are therefore excluded. $\endgroup$ Mar 30, 2011 at 3:31
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    $\begingroup$ Note that the "reverse" reaction would have to be something like (SM stuff) + (dark matter) -> (other SM stuff) + (charged SUSY particle). The cross-section for a reaction with only one product is always going to vanish because the reactants have to have exactly the right kinetic energy to conserve energy and momentum (or equivalently the phase space of that single product particle, which goes into the cross section, vanishes). But with that correction it seems like a very good question. $\endgroup$ Mar 30, 2011 at 3:35
  • $\begingroup$ @Keenan Oops! Good point. I changed the text per your suggestion. $\endgroup$
    – John
    Mar 30, 2011 at 3:51
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    $\begingroup$ @Jerry I'm already assuming the lightest SUSY particle is uncharged, so your comment doesn't make much sense. Either you misunderstood the question, or I'm misunderstanding your comment. $\endgroup$
    – John
    Mar 30, 2011 at 3:53

1 Answer 1


Dear John, your basic logic is valid but the way you "estimate" the cross section for the LSP production by the time reversal of the decay is way too emotional.

The reaction by which you expect charged dark matter to be created everywhere, namely $${\rm SM stuff} + {\rm LSP}_{\rm from\,dark\,matter} \to {\rm charged\, superpartners}$$ is indeed possible and it is essentially what the underground detectors of dark matter are trying to find. However, this reaction is extremely rare because dark matter, while everywhere, is very diluted - less than half a GeV per cubic centimeter


With this - relatively to nuclear densities - tiny mass density (even smaller than the cosmological constant), it takes some time to detect the collisions with the dark matter.

You may also misapply the time reversal. While the laws of physics produce CPT-invariant - and approximately T-invariant - Lorentz-invariant amplitudes, the actual probabilities of the opposite processes contain different kinematic factors: that's implied by the time asymmetry of mathematical logic, an elementary fact that - as I found out - a huge number of people misunderstand.

In particular, the decay of the charged superpartner to the LSP and a charged standard model particle is described by the "decay rate" while the opposite process is described by a "cross section". The cross section and the decay rate don't even have the same dimension (units). So it's obvious that they can't be "equal". And saying that one of them is "small" if the other is "small" is only true if you properly "normalize" what you mean by the vague adjective "small". You apparently didn't do it right. The sentence is surely not right if "small" is meant as "small for purposes I find practical".

The "decay rate" and the "cross section" of the two processes related by the time reversal differ by kinematic factors comparable to powers of the total energy of a particle in the process. If the decay rate (width) is e.g. $1 {\rm GeV}$, which makes the particle decay almost immediately in the detector, and if its mass is $100 {\rm GeV}$, the cross section, the corresponding cross section might be comparable to $1 {\rm GeV}/ (100 {\rm GeV})^3$ which is a tiny cross section. It was made tiny by the huge denominator, and the denominator is huge because the mass of the superpartner is high.

The two processes differing by "time reversal" have different prior probabilities. There is simply no "simple" way to relate the opposite probabilities. In a similar way, if glass "certainly" breaks down if it falls to the floor, it doesn't mean that it's likely that the pieces will "often" collect themselves and create a perfect unbroken glass. The opposite process is suppressed, relatively to the first one, by $\exp(-\Delta S)$, the exponential of the entropy difference between the initial and final state.

But even for microscopic processes, the relevant "probabilities" of the process have - aside from the shared $|{\mathcal M}|^2$, the squared Lorentz-invariant amplitude, lots of kinematic factors - namely $1/2E$ from both initial and final particles (which is symmetric with respect to the time reversal); and the integration over $d^3 p$ of final particles (not initial ones, which is time-reversal-asymmetric). So expecting that the probabilities of the time-reversed evolutions are "equal" is a trivial logical fallacy. They're not equal - that's not what T-invariance or CPT-invariance mean.

  • $\begingroup$ My understanding was that underground WIMP detectors are looking for scattering events, not charged SUSY creation. Anyway, thank you! It is posters like you that really make this site great, by taking the time to go beyond pointing out a mistake to explaining why it is a mistake and how to correct it, the answer becomes very useful. Much appreciated! $\endgroup$
    – John
    Mar 30, 2011 at 18:43

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